`A(g) + 3B(g) rarr 4C(g)`Initially concentration of A is equal to that of B. The equilibrium concentrations of A and C are equal. Kc is :

Options

A. 0.08
B. 8
C. 80
D. `1//8`

Explanation

To solve the problem step by step, we will analyze the given chemical equilibrium and derive the equilibrium constant \( K_c \).

**Step 1: Write the balanced chemical equation.**
The reaction is given as:
\[ A(g) + 3B(g) \rightleftharpoons 4C(g) \]

**Step 2: Define initial concentrations.**
Let the initial concentrations of \( A \) and \( B \) be \( a \) (since they are equal). The initial concentration of \( C \) is zero:
- \([A]_0 = a\)
- \([B]_0 = a\)
- \([C]_0 = 0\)

**Step 3: Set up the changes in concentrations at equilibrium.**
Let \( x \) be the amount of \( A \) that reacts at equilibrium. The changes in concentrations will be:
- For \( A \): \( [A] = a - x \)
- For \( B \): \( [B] = a - 3x \) (since 3 moles of \( B \) react for every mole of \( A \))
- For \( C \): \( [C] = 4x \) (since 4 moles of \( C \) are produced for every mole of \( A \))

**Step 4: Use the information given about equilibrium concentrations.**
We know that the equilibrium concentrations of \( A \) and \( C \) are equal:
\[ a - x = 4x \]

**Step 5: Solve for \( x \).**
Rearranging the equation:
\[ a = 5x \]
Thus, we can express \( x \) in terms of \( a \):
\[ x = \frac{a}{5} \]

**Step 6: Substitute \( x \) back to find equilibrium concentrations.**
Now we can find the equilibrium concentrations:
- For \( A \):
\[ [A] = a - x = a - \frac{a}{5} = \frac{4a}{5} \]
- For \( B \):
\[ [B] = a - 3x = a - 3\left(\frac{a}{5}\right) = a - \frac{3a}{5} = \frac{2a}{5} \]
- For \( C \):
\[ [C] = 4x = 4\left(\frac{a}{5}\right) = \frac{4a}{5} \]

**Step 7: Write the expression for the equilibrium constant \( K_c \).**
The expression for \( K_c \) is given by:
\[ K_c = \frac{[C]^4}{[A]^1[B]^3} \]

**Step 8: Substitute the equilibrium concentrations into the \( K_c \) expression.**
Substituting the values we found:
\[ K_c = \frac{\left(\frac{4a}{5}\right)^4}{\left(\frac{4a}{5}\right)^1 \left(\frac{2a}{5}\right)^3} \]

**Step 9: Simplify the expression.**
Calculating the numerator:
\[ \left(\frac{4a}{5}\right)^4 = \frac{256a^4}{625} \]
Calculating the denominator:
\[ \left(\frac{4a}{5}\right) \left(\frac{2a}{5}\right)^3 = \frac{4a}{5} \cdot \frac{8a^3}{125} = \frac{32a^4}{625} \]

Now substituting back into the \( K_c \) expression:
\[ K_c = \frac{\frac{256a^4}{625}}{\frac{32a^4}{625}} = \frac{256}{32} = 8 \]

**Step 10: Conclusion.**
Thus, the equilibrium constant \( K_c \) is:
\[ K_c = 8 \]